The vapour pressure of pure liquid A at 300 K is 577 Torr and that of pure liquid B is 390 Torr. These two compounds form ideal liquid and gaseous mixtures. Consider the equilibrium composition of a mixture in which the mole fraction of A in the vapour is 0.35. Find the mole % of A in liquid.
Text Solution
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(27)
A and B are volatile liquids, given P A 0 = 575 Torr , P B 0 = 390 Torr
let mole fraction of A in solution = X A
hence, P total = P A 0 X A + P B 0 (1 – X A )
also X A ′ = mole fraction of A in the vapour = 0.35
X A ′ = 
= 0.35
= 
this gives X A = 0.27
Composition of liquid mixture, A = 27 mol % , B = 73 mol %.
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